Q. Starting with 1 mole of $N_{2}O_{4},\$if $\alpha$ is the degree of dissociation of $N_{2}O_{4}$ for the reaction, $N_{2}O_{4} \rightleftharpoons 2{NO}_{2}$ then at equilibrium the total number of moles of $N_{2}O_{4}$ and $NO_{2}$ present is
A) $2$
B) $(1 - \alpha)$
C) $(1 - \alpha)^{2}$
D) $(1 + \alpha)$
Explanation
(d)
$N_{2}O_{4} \rightleftharpoons 2{NO}_{2}$
$(1 - \alpha)\ \ \ \ 2\alpha$
Total number of moles $= 1 - \alpha + 2\alpha = 1 + \alpha$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.